The opposite triangle is $(30 – 60 -90)$ triangle which side lengths are with the ratio $x$, $x sqrt {3}$ and $2x$

Since $overline {AC}=1$          (Given)

Then:

$overline {BC}=dfrac {1}{2}$

$overline {AB}=dfrac {sqrt {3}}{2}$

The opposite triangle is $(45 – 45 -90)$ triangle which side lengths are with the ratio $x$, $x$ and $x sqrt {2}$

Since $overline {AC}=1$          (Given)

Then:

$overline {BC}=dfrac {1}{sqrt {2}}$

$overline {AB}=dfrac {1}{sqrt {2}}$

The missing side lengths of the $(30 – 60 -90)$ triangle are:          $dfrac {1}{2}$ and $dfrac {sqrt {3}}{2}$

The missing side lengths of the $(45 – 45 -90)$ triangle are:          $dfrac {1}{sqrt {2}}$ and $dfrac {1}{sqrt {2}}$

The sum of logarithms is the logarithm of the product and the difference of the logarithm is the logarithm of the quotient:

a. $log_3{5m}$

b. $log_6{dfrac{p}{m}}$

c. $log_2{r}+3log_6{z}=log_2{r}+3dfrac{log_2{z}}{log_2{6}}=log_2{r}+dfrac{log_2{z}}{dfrac{1}{3}log_2{6}}=log_2{r}+dfrac{log_2{z}}{log_2{6^{1/3}}}=log_2{r}+dfrac{log_2{z}}{log_2{sqrt[3]{6}}}=log_2{r}+log_2{z^{1/log_2{sqrt[3]{6}}}}=log_2{rz^{1/log_2{sqrt[3]{6}}}}$

d. $log{dfrac{90cdot 4}{36}}=log{10}=1$

a. $log_3{5m}$

b. $log_6{dfrac{p}{m}}$

c. $log_2{rz^{log_2{1/sqrt[3]{6}}}}$

d. $1$

In this exercise we can use the following formula

$$
color{#c34632}{(a+b)^2 = a^2 + 2ab + b^2}
$$

Now, we have

$$
begin{align*}
f(x) &= (x+4)(x+1)^2 (x-2)\
&= (x+4)(x^2 + 2x +1)(x-2)\
&= (x^3 + 2x^2 +x+4x^2 +8x +4) (x-2)\
&= (x^3 + 6x^2 +9x +4) (x-2)\
&= x^4 + 6x^3 +9x^2 +4x – 2x^3 – 12x^2 -18x-8\
&= x^4 + 4x^3 – 3x^2 – 14x-8
end{align*}
$$

This function is a fourth-degree.

$$
f(x) = x^4 + 4x^3 – 3x^2 – 14x-8
$$

$$
CBD=A+ACB=90text{textdegree}+20text{textdegree}=110text{textdegree}
$$

We determine the angle $CBD$:

$BCD=180text{textdegree}-CBD-D$

$=180text{textdegree}-110text{textdegree}-68text{textdegree}$

$$
=1text{textdegree}
$$

We determine the angle BCD$:

$dfrac{sin BCD}{BD}=dfrac{sin D}{BC}$

$sin BCDcdot BC=BDsin D$

$BC=dfrac{125cdot sin 1text{textdegree}}{sin 68text{textdegree}}$

$=dfrac{125cdot 0.01745241}{0.92718385}$

$$
approx 2.35
$$

We determine the side $BD$, using the Law of Sines:

$2.35$ feet

The next four terms are $t(3)=15,t(4)=21,t(5)=27,t(6)=33,$

(a) For an arithmetic sequence, each term is given by the sum of the previous term and the common difference or such that the n’th term can be expressed in terms of the first term and a summation of common differences up to the n’th term. Here, we have $d=9-3=6$ such that $t(n)=t(1)+(n-1)d$ eg t(6)=3+5d=33

The next four terms are $t(3)=27,t(4)=81,t(5)=243,t(6)=729,$

(b) For a geometric sequence, each term is given by multiplying the previous term by a constant factor r where r is the ratio of successive elements or such that the n’th term can be expressed in terms of the first term factored by a power of r. Here, we have $r=dfrac{9}{3}=3$ such that $t(n)=t(1)r^{n-1}$ eg $t(6)=3times3^{5}=729$

The next four terms are $t(2)=-9,t(3)=27,t(4)=-81,t(5)=243,$

(c) Define a series with alternating positive and negative terms by $t(n)=(-1)^{n+1}t(1)r^{n-1}$

$$
n^3=49
$$

$$
n=sqrt[3]{49}approx 3.66
$$

$$
3^n=49
$$

$log (3^n)=log (49)$

$nlog (3)=log (49)$

$$
n=dfrac{log (49)}{log (3)}approx 3.54
$$

$$
3^{n+1}=49
$$

$log (3^{n+1})=log (49)$

$(n+1)log (3)=log (49)$

$n+1=dfrac{log (49)}{log (3)}$

$$
n=dfrac{log (49)}{log (3)}-1approx 2.54
$$

a) $sqrt[3]{49}approx 3.66$

b) $dfrac{log (49)}{log (3)}approx 3.54$

c) $dfrac{log (49)}{log (3)}-1approx 2.54$

c) We notice that the function grows as the angle of rotation grows from $0text{textdegree}$ to $90text{textdegree}$ and from $270text{textdegree}$ to $360text{textdegree}$ and decreases on the other interval. So we can divide portions of intervals and use an average value.

$sin 40text{textdegree}=dfrac{h}{1}$

$$
h=sin 40text{textdegree}
$$

We determine an equation for $h$:

The height $h$ is similar to the heights we computed in Problem 9-10.

$$
y=sintheta
$$

b) We write an equation for the escape height $y$ for any passenger, given the rotation angle $theta$:

d) We graph the equation for larger values of $theta$:

We notice that the function repeats itself each $360text{textdegree}$, but this makes sense because the position of the seat in the Screamer wheel repeats after each rotation.

begin{center}
begin{tabular}{|| c|c|c| c||}
hline
Degree of rotation,$x$ & Height & Actual height & $y=sin x$ \ [0.5ex]
hline
0&0&0 & 0\
hline
30& 0.5 & 50 & 0.50\
hline
45& 0.7 & 71 & 0.71\
hline
60& 0.9 & 87 & 0.87\
hline
90& 1 & 100 & 1\
hline
120& 0.9 & 87 & 0.87\
hline
135& 0.7 & 71 & 0.71\
hline
150& 0.5 & 50& 0.50\
hline
180& 0 & 0 & 0\
hline
210& -0.5 & -50 & -0.50\
hline
225& -0.7 & -71 & -0.71\
hline
240& -0.9 & -87 & -0.87\
hline
270& -1 & -100 & -1\
hline
300& -0.9 & -87 & -0.87\
hline
315& -0.7 & -71 & -0.71\
hline
330& -0.5 & -50 & -0.50\
hline
360& 0 & 0 & 0\[1ex]
hline
end{tabular}
end{center}

d) The angle of rotation from the circle are the $x$-coordinates in the graph. The heights of the triangles built in the circle are the $y$-coordinates from the graph.

a-

$30text{textdegree} – 60text{textdegree} – 90text{textdegree}$ triangle sides are with the ratio $1: sqrt {3}:2$

$45text{textdegree} – 45text{textdegree} – 90text{textdegree}$ triangle sides are with the ratio $1: 1: sqrt {2}$

b-

A $30text{textdegree} – 60text{textdegree} – 90text{textdegree}$ triangle is called a half-equilateral because when reflecting it around the long leg we obtain an equilateral.

a-          $30text{textdegree} – 60text{textdegree} – 90text{textdegree}$ triangle sides are with the ratio $1: sqrt {3}:2$

.          $45text{textdegree} – 45text{textdegree} – 90text{textdegree}$ triangle sides are with the ratio $1: 1: sqrt {2}$

b-          A $30text{textdegree} – 60text{textdegree} – 90text{textdegree}$ triangle is called a half-equilateral because when reflecting it around the long leg we obtain an equilateral.

$sin A=dfrac {0.3}{1}$

$mangle A=17.46$$text{textdegree}$

$mangle A=17.46$$text{textdegree}$

$$
y=a(x-h)^2+k
$$

$k=15$

$h=dfrac{250}{2}=125$

$$
Rightarrow y=a(x-125)^2+15
$$

We determine $h, k$:

$0=a(0-125)^2+15$

$0=15,625a+15$

$$
a=-dfrac{15}{15,625}=-0.00096
$$

We determine $a$ using the point $(0,0)$:

$$
y=-0.00096(x-125)^2+15
$$

$h=dfrac{240}{2}=120$

$$
y=a(x-120)^2+k
$$

b) The ball of Dwayne’s brother traveled $220+20=240$ meters. Therefore we have:

$$
begin{cases}
a(0-120)^2+k=0\
a(20-120)^2+k=5.5
end{cases}
$$

$$
begin{cases}
14,400a+k=0\
10,000a+k=5.5
end{cases}
$$

$14,400a+k-10,000a-k=0-5.5$

$4400a=-5.5$

$a=-dfrac{5.5}{4400}=-0.00125$

$$
k=-14,400(-0.00125)=18
$$

We determine $a, k$ using two points from the graph:

$$
y=-0.00125(x-120)^2+18
$$

Therefore David’s ball traveled farther horizontally ($250>240$), but Dwayne’s ball traveled higher ($18>15$).

The expression

$$
x^3 – 2x^2 + 25x -50
$$

we can write

$$
begin{align*}
x^3 – 2x^2 + 25x – 50 &= x^2 (x-2)+25 (x-2)\
&= (x-2)(x^2 + 25)
end{align*}
$$

Now, we have

$$
begin{align*}
frac {x^3 – 2x^2 + 25x – 50}{x-2} &= frac {(x-2)(x^2 +25)}{x-2}\
&= x^2 + 25
end{align*}
$$

Therefore

$$
frac {x^3 – 2x^2 + 25x – 50}{x-2} = x^2 + 25
$$

$$
frac {x^3 – 2x^2 + 25x – 50}{x-2} = x^2 + 25
$$

Since

$$
x^3 – 2x^2 + 25x – 50 = (x-2)(x^2 + 25)
$$

we have

$$
begin{align*}
x^3 – 2x^2 + 25x – 50 &= 0\
(x-2)(x^2 + 25) &= 0\
x-2 = 0 qquad x^2 + 25 &= 0\
x=2 qquad x^2 &= -25\
x &= pm sqrt {-25}\
x &= pm 5_i
end{align*}
$$

The solutions of equation are

$$
x=2 qquad x=5_i qquad x=-5_i
$$

$$
x=2 qquad x=5_i qquad x=-5_i
$$

$$
m_{2-6}=dfrac{160-700}{6-2}=-dfrac{540}{4}=-135
$$

$$
m_{10-13}=dfrac{80-50}{13-10}=dfrac{30}{3}=10
$$

$r=2$ feet$=2cdot 12=24$ inches

$V=dfrac{4pi r^3}{3}=dfrac{4picdot 24^3}{3}approx 57,906$ cubic inches

$V=dfrac{4pi r^3}{3}cdot 12^3=dfrac{4cdot 12^3}{3}cdot picdot r^3$

$$
=2304pi r^3
$$

a) $57,906$ cubic inches

b) $V=2304pi r^3$

The domain of the function

$$
y = sin theta
$$

is

$$
( – infty , + infty )
$$

$$
( – infty , + infty )
$$

$sin AOB=dfrac{AB}{OA}=dfrac{65}{100}=0.65$

$sin^{-1} 0,65approx 40.5text{textdegree}$

$AOB=40.5text{textdegree}$ or $AOB=180text{textdegree}-40.5text{textdegree}=139.5text{textdegree}$

We determine the angle $AOB$ using the sine function:

$$
40.5text{textdegree}; 139.5text{textdegree}
$$

If the function equation contains a factor $(x-a)^2$ then $x=a$ is a double root:

$t(n)=50-7n$

$h(n)=4cdot 3^n$

$$
q(n)=n^2-6n+17
$$

$t(n+1)-t(n)=[50-7(n+1)]-(50-7n)$

$=50-7n-7-50+7n$

$$
-7
$$

a) We check if the sequence $t(n)$ is arithmetic or geometric:

$a_1=50-7(1)=43$

$$
d=-7
$$

The sequence $t(n)$ is arithmetic:

$$
dfrac{h(n+1)}{h(n)}=dfrac{4cdot 3^{n+1}}{4cdot 3^n}=3
$$

We check if the sequence $h(n)$ is arithmetic or geometric:

$h(1)=4cdot 3^1=12$

$$
q=3
$$

The sequence $h(n)$ is geometric:

$q(n+1)-q(n)=(n+1)^2-6(n+1)+17-n^2+6n-17$

$=n^2+2n+1-6n-6+17-n^2+6n-17$

$=2n+1$

$dfrac{q(n+1)}{q(n)}=dfrac{(n+1)^2-6(n+1)+17}{n^2-6n+17}$

$q(n)=(n^2+6n+9)+8=(n+3)^2+8$

$$
q(1)=(1+3)^2+9=16+9=25
$$

We check if the sequence $q(n)$ is arithmetic or geometric:

The sequence $q(n)$ is neither arithmetic, nor geometric.

$$
43, 36, 29, 22, 15, 8….
$$

b) The first sequence is decreasing, while the second is increasing. The last sequence in decreasing for $nin {1, 2, 3}$ and increasing for $n>3$. We write the first terms of $t(n)$:

This means the sequence $t(n)$ has no term in common with the other two,therefore there is no common term for the three sequences..

$n=1$

$h(1)=12; h(2)=36; h(3)=108$

$q(1)=12$

$q(2)=9$

$q(3)=8$

$q(4)=9$

$$
q(5)=72
$$

c) The exponential grows faster than the quadratic from some point further. The only possibility that the two sequences share a term is among the small values of $n$:

$$
h(1)=t(1)=12
$$

$$
2^{x-1}=64
$$

$2^{x-1}=2^6$

$x-1=6$

$x=1+6$

$$
x=7
$$

$$
9^3=17^{2x-1}
$$

$(3^2)^3=(3^3)^{2x-1}$

$3^{2cdot 3}=3^{3(2x-1)}$

$3^6=3^{6x-3}$

$6=6x-3$

$6x=6+3$

$6x=9$

$x=dfrac{9}{6}$

$$
x=dfrac{3}{2}
$$

$x^6=29$

$x=sqrt[6]{29}approx 1.75$

$6^x=29$

$ln 6^x=ln 29$

$xln 6=ln 29$

$x=dfrac{ln 29}{ln 6}$

$$
xapprox 1.88
$$

a) $x=7$

b) $x=dfrac{3}{2}$

c) $xapprox 1.75$
d) $xapprox 1.88$

a) In order to calculate $x$, we can use the sine function

$$
sin theta = frac ax
$$

where $a= frac 12$ and $theta = 30^{text{textdegree}}$. Then

$$
begin{align*}
sin 30^{text{textdegree}} &= frac {frac 12}{x}\
frac 12 &= frac {1}{2x}\
x cdot frac 12 &= frac 12\
x &= 1
end{align*}
$$

b) In order to calculate $x$, we can use the cosine function

$$
cos theta = frac xa
$$

where $a=1$ and $theta = 45^{text{textdegree}}$. Then

$$
begin{align*}
cos 45^{text{textdegree}} &= frac x1\
frac {sqrt 2}{2} &= x
end{align*}
$$

a) $x=1$

b) $frac {sqrt 2}{2} = x$

Considering the circular rugs have 5 feet diameter, it means on a horizontal side of the floor she needs $dfrac{20}{5}=8$ rugs, while on the other side she would need $dfrac{20}{5}=4$ rugs, which gives a total of $5cdot 4=20$ rugs.

$A_{rugs}=20cdotpicdot left(dfrac{5}{2}right)^2approx392.7$ square feet

$A_{floor}=25cdot 20=500$ square feet

$$
dfrac{A_{rugs}}{A_{floor}}=dfrac{392.7}{500}=0.7854=78.54%
$$

Since $78.54<80$, she shouldn't buy the rugs as they don't cover $980%$ from the floor.

b) There are another 3 seats from which the distance to the ground in the same: $C, D,E$:

$AOD=180text{textdegree}-theta$

$AOB=180text{textdegree}+theta$

$AOE=360text{textdegree}-theta$.

c) $triangle OCNcongtriangle ODMcongtriangle OBMcongtriangle QEN$.

Let’s suppose we are given

$theta=AOC$.

We express the angles $AOD, AOB, AOE$ in terms of $theta$:

$$
sin 130text{textdegree}approx 0.77
$$

This means the height of the chair rotated by $130text{textdegree}$ is 77 feet.

b) There is another seat at the same height:

$$
180text{textdegree}-130text{textdegree}=50text{textdegree}
$$

c) We can use the graph to determine angles for which the $y$-coordinates are the same: we plot the point corresponding to $x=130text{textdegree}$, then we draw a parallel to the $x$-axis. The point(s) in which the line intersects the graph give us the solution(s):

$$
theta=80text{textdegree}
$$

$sintheta=sin 80text{textdegree}approx 0.98=100text{textdegree}$

$$
theta=200text{textdegree}
$$

$$
sintheta=sin 200text{textdegree}approx -0.34=sin 340text{textdegree}
$$

$$
theta=310text{textdegree}
$$

$$
sintheta=sin 310text{textdegree}approx -0.77=sin 230text{textdegree}
$$

We can actually define $sin x$ and graph the function $y=sin x$ by using the unit circle.

The point $P_x$ of the unit circle. It lies on the line that closes an angle that measures $x^{circ}$ with the $x$-axis.

The $y$-coordinate of the point $P_x$ (the height from the $x$-axis to the unit circle when the angle is $x^{circ}$) is actually $sin x$.

By finding the heights (distances from the unit circle) we find the values of the function $sin x$ for various measures $x^{circ}$.

We sketch the first two cycles of $y=sin(theta)$.

We draw a unit circle and mark each of the point $A, B, C$ on it:

In $A$ the rider has just passed the highest point, in $B$ the rider has just starting the descent from the ground level to the water pit,while in $C$ the rider got back to the initial point on the ground.

Given equations :
$$y = log_2(x-1)rightarrow(1)$$
$$y = x^3 – 4xrightarrow(2)$$
Intersecting point : $(2, 0)$ and $(1.1187, -3.075)$.

Putting, x = 2 in $log_2(x-1) = x^3 – 4x$.
$$begin{aligned}
log_2(2-1)&= (2)^3 – 4(2)\
log_2(1)&= 8 – 8\
log_2(1)&= 0 \
0 = 0\

end{aligned}$$

a) The expression

$$
begin{align*}
x^5 + 8x^2 y^3 &= x^2 ( x^3 + 8y^3)\
&= x^2 [ x^3 + (2y)^3]
end{align*}
$$

According to the formula

$$
begin{align*}
color{#c34632}{a^3 + b^3 = (a+b)(a^2 – 2ab + b^2)} tag {*}
end{align*}
$$

we obtain

$$
x^5 + 8x^2 y^3 = x^2 (x+2y)(x^2 – 2xy +4y^2)
$$

b) The expression

$$
8y^6 – 125x^3 = (2y^2)^3 – (5x)^3
$$

According to the $(*)$ we have

$$
8y^6 – 125^3 = (2y^2 + 5x)(4y^4 – 10xy^2 + 25x^2)
$$

a) $x^5 + 8x^2 y^3 = x^2 (x+2y)(x^2 – 2xy +4y^2)$

b) $8y^6 – 125^3 = (2y^2 + 5x)(4y^4 – 10xy^2 + 25x^2)$

In this exercise we can use the following formula

$$
color{#c34632}{c^2 = a^2 + b^2 – 2ab cos theta}
$$

where

$$
begin{align*}
a &= 620 text{ m }\
b &= 455 text{ m }\
theta &= 150^{text{textdegree}}
end{align*}
$$

Then we have

$$
begin{align*}
c^2 &= 620^2 + 455^2 – 2 cdot 620 cdot 450 cos 150^{text{textdegree}}\
c^2 &= 384400 + 207025 – 564200 cdot (-0.86)\
c^2 &= 591425 + 485 212\
c^2 &= 1076637\
c &= sqrt {1076637}\
c &= 1037.61
end{align*}
$$

The lake is $1037.61$ meters long.

Yee Ping can’t swim the lake.

$$
c = 1037.61
$$

Let

$$
begin{align*}
a &= 2 +3 i\
b &= 1- i
end{align*}
$$

a) We will add $a$ and $b$

$$
begin{align*}
a+b &= (2+ 3 i) + (1- i)\
&= 2 + 3 i + 1 – i\
&= 3 + i (3 -1)\
&= 3 + 2 i
end{align*}
$$

b) We subtract $a$ and $b$

$$
begin{align*}
a-b &= (2+ 3 i) – (1- i)\
&= 2 +3 i – 1 + i\
&= 1 + i (3+1)\
&= 1 + 4 i
end{align*}
$$

c) We will multiply $a$ and $b$

$$
begin{align*}
a cdot b &= (2+3 i) (1- i)\
&= 2 – 2 i + 3 i – 3 i^2\
&= 2 + i – 3 i^2
end{align*}
$$

Since

$$
i^2 = -1
$$

we obtain

$$
begin{align*}
a cdot b &= 2 + i – 3 cdot (-1)\
&= 2 + i + 3\
&= 5 + i
end{align*}
$$

a) $a+b = 3 + 2 i$

b) $a-b = 1 + 4_i$

c) $a cdot b = 5 + i$

$V=(2r)^3=(2cdot 2)^3=64$ cubic feet

$V=(2r)^3$

$13=8r^3$

$r^3=dfrac{13}{8}$

$r=sqrt[3]{dfrac{13}{8}}approx 1.176$ feet

$v=(2r)^3$

$v=8r^3$

$$
r=dfrac{sqrt[3]{v}}{2}
$$

c) We determine the radius of the largest spherical pinata that a box with volume $v$ can hold:

a) 4 ft, 4ft, 4ft, 64 cubic feet

b) 1.176 ft

c) $r=dfrac{sqrt[3]{v}}{2}$

$x=OR$

$$
y=PR
$$

We draw $PRperp Ox$.

The coordinates $(x,y)$ of the point $P$ are:

$cos 34text{textdegree}=dfrac{OR}{OP}$

$OR=OPcos 34text{textdegree}$

$OR=1cdot cos 34text{textdegree}$

$$
textcolor{#4257b2}{OR=cos 34text{textdegree}}
$$

We use the cosine function in $triangle OPR$:

$sin 34text{textdegree}=dfrac{PR}{OP}$

$PR=OPsin 34text{textdegree}$

$PR=1cdot sin 34text{textdegree}$

$$
textcolor{#4257b2}{PR=sin 34text{textdegree}}
$$

We use the sine function in $triangle OPR$:

$$
P(cos 34text{textdegree},sin 34text{textdegree})
$$

The coordinates of the point $P$ are:

$$
P(cos 34text{textdegree},sin 34text{textdegree})
$$

$x=OR$

$$
y=QR
$$

We draw $QRperp Ox$.

The coordinates $(x,y)$ of the point $Q$ are:

$cos thetatext{textdegree}=dfrac{OR}{OQ}$

$OR=OQcos theta$

$OR=1cdot cos theta$

$$
textcolor{#4257b2}{OR=cos theta}
$$

We use the cosine function in $triangle OQR$:

$sin theta=dfrac{QR}{OQ}$

$QR=OPsin theta$

$QR=1cdot sin theta$

$$
textcolor{#4257b2}{QR=sin theta}
$$

We use the sine function in $triangle OQR$:

$$
Q(cos theta,sin theta)
$$

The coordinates of the point $Q$ are:

$$
Q(cos theta,sin theta)
$$

If there is a point $B$ on a circle whereas the segment connecting $B$ with the origin is making an angle $theta$ with the x-axis,

Then $sin theta$ tells us the ratio between y-coordinate of $B$ and the radios of the circle.

$cos theta$ tells us the ratio between x-coordinate of $B$ and the radios of the circle

$sin theta$ tells us the ratio between y-coordinate of $B$ and the radios of the circle.

$cos theta$ tells us the ratio between x-coordinate of $B$ and the radios of the circle

If we know the sine of an angle in a unit circle, we can find its cosine.

The opposite graph is a unit circle with the point $B$ on the circle.

Since $sin theta$ is the y-coordinate of $B$

$cos theta$ is the x-coordinate of $B$

$1^2=x^2+y^2$          (Pythagorean theorem)

$x=sqrt {1^2-y^2}$

$cos theta=sqrt {1^2-y^2}$

$cos theta=sqrt {1^2-(sin theta)^2}$

$cos theta=sqrt {1^2-(sin theta)^2}$

a-

Since, $cos theta=$ x-coordinate

Then x-coordinate of the point on the unit circle is $dfrac {3}{4}$

$1^2=x^2+y^2$          (Pythagorean theorem)

$y=sqrt {1-x^2}$

$y=sqrt {1-dfrac {9}{16}}=sqrt {dfrac {7}{16}}$

$y=dfrac {sqrt 7}{4}$

The exact coordinates of point $Q$ are $(dfrac {3}{4}, qquad dfrac {sqrt 7}{4})$

b-

$sin theta=$ y-coordinate of the point $Q$

$$
sin theta=dfrac {sqrt 7}{4}
$$

c-

Since $(x, y)$ are the coordinates of the point $Q$ on the unit circle, then:

$1^2=x^2+y^2$          (Pythagorean theorem)

$sin theta=y$

$cos theta=x$

So,

$$
1^2=(sin theta)^2+(cos theta)^2
$$

a-          The exact coordinates of point $Q$ are $(dfrac {3}{4}, qquad dfrac {sqrt 7}{4})$

b-          $sin theta=dfrac {sqrt 7}{4}$

c-          $1^2=(sin theta)^2+(cos theta)^2$

The idea is that the length of the base of a triangle is the value of $cos x$, where $x$ is the angle of a triangle next to the base.

Let us draw $20$ triangles in a unit circle and plot the points with the angles as $x$-coordinates and the lengths of the bases of the triangles as the $y$-coordinates.

The graphs of $sin x$ and $cos x$ are of the same shape, but the graph of $cos x$ is shifted $dfrac{pi}{2}$ to the right (or left, it does not matter since they are periodic).

$cos AOB=dfrac{OB}{OA}=dfrac{27}{100}=0.27$

$cos^{-1} 0.27approx 74.3text{textdegree}$

$$
AOB=74.3text{textdegree}
$$

There are 4 possibilities: $A, C, E, F$. First we determine the angle $AOB$, using cosine:

$COB=180text{textdegree}-74.3text{textdegree}=105.7text{textdegree}$

$EOB=180text{textdegree}+74.3text{textdegree}=254.3text{textdegree}$

$$
FOB=360text{textdegree}-74.3text{textdegree}=285.7text{textdegree}
$$

Then we determine all the 3 angles in the other quadrants which have the same reference angle of $58text{textdegree}$:

$$
74.3text{textdegree}; 105.7text{textdegree}; 254.3text{textdegree}; 285.7text{textdegree}
$$

a-

The following is the graph of the function $y=cos(theta)$

From the graph:

$cos (130text{textdegree})=-0.6428$

b-

From the graph, the other angel that has the same base is $230$$text{textdegree}$

c-

We can use the symmetry of the cosine-calculator graph to calculate the angle location by drawing the line $x=180$ and reflect the point $(130, -0.6428)$ about it to get the corresponding point as in the graph below.

a-          $cos (130text{textdegree})=-0.6428$

b-          The other angel that has the same base is $230$$text{textdegree}$

The point $P$ has an angle of $50$$text{textdegree}$

$sin 50=0.766$

$cos 50=0.6428$

The coordinates of $P$ are:          $(0.6428, qquad 0.766)$

The point $Q$ has an angle of $110$$text{textdegree}$

$sin 110=0.9397$

$cos 110=-0.342$

The coordinates of $Q$ are:          $(-0.342, qquad 0.9397)$

The coordinates of $P$ are:          $(0.6428, qquad 0.766)$

The coordinates of $Q$ are:          $(-0.342, qquad 0.9397)$

a-

$mangle ROS=60$$text{textdegree}$          (Given)

$overline {OR}$ rotation angle $=360-60=300$$text{textdegree}$

b-

$cos {SOR}=dfrac {overline {OS}}{overline {OR}}=dfrac {overline {OS}}{1}=overline {OS}$

$cos 60=0.5$

$overline {OS}=0.5$ units

$sin {SOR}=dfrac {overline {SR}}{overline {OR}}=dfrac {overline {SR}}{1}=overline {SR}$

$sin 60=0.866$

$overline {SR}=0.866$ units

c-

The exact coordinates of point $R$ are:          $(0.5, qquad -0.866)$

a-          $overline {OR}$ rotation angle $=360-60=300$$text{textdegree}$

b-          $overline {OS}=0.5$ units          $overline {SR}=0.866$ units

c-          The exact coordinates of point $R$ are:          $(0.5, qquad -0.866)$

$sin theta=dfrac {1}{4}$          (Given)

$(sin theta)^2+(cos theta)^2=1$          (Pythagorean Identity)

$(cos theta)^2=1-left(dfrac {1}{4} right)^2$

$(cos theta)^2=dfrac {16}{16}-dfrac {1}{16}$

$(cos theta)^2=dfrac {15}{16}$

$cos theta=pm dfrac {sqrt {15}}{4}$

The exact coordinates of a point on the unit circle than has $sin theta=dfrac {1}{4}$ are:

$left(dfrac {sqrt {15}}{4}, qquad dfrac {1}{4} right)$

$$
left(dfrac {sqrt {15}}{4}, qquad dfrac {1}{4} right)
$$

$P=1000$

$r=0.06$

$A=4000$

$$
n=12
$$

$4000=1000left(1+dfrac{0.06}{12}right)^{12t}$

In order to determine the number of years $t$ until the account is worth $4000$ we use the formula:

$A=Pleft(1+dfrac{r}{n}right)^{nt}$.

$dfrac{4000}{1000}=left(1+dfrac{0.06}{12}right)^{12t}$

$4=left(1+dfrac{0.06}{12}right)^{12t}$

$$
4=(1.005)^{12t}
$$

$ln 4=ln left(1.005right)^{12t}$

$ln 4=12tln 1.005$

$t=dfrac{ln 4}{12ln 1.005}$

$$
tapprox 23.16
$$

We divide both sides by $12ln 1.005$:

$tapprox 23.16$ years

a. The possible factors are of the form $(x-a)$ and $(x+a)$ with $a$ divisors of the constant term of the polynomial:

$$
(x-1),(x+1),(x-7),(x+7)
$$

b.
$$
polylongdiv{x^4-6x^3-6x^2+6x-7}{x+1}
$$

Since the remainder is nonzero, $(x+1)$ is not a factor.

$$
polylongdiv{x^4-6x^3-6x^2+6x-7}{x-1}
$$

Since the remainder is nonzero, $(x-1)$ is not a factor.

a. $(x-1),(x+1),(x-7),(x+7)$

b. Neither are factors

Let

$$
begin{align*}
f(x) &= 2x^2 + 5x -3\
g(x) &= x^2 + 4x +3
end{align*}
$$

a) Therefore

$$
begin{align*}
f(x) &= g(x)\
2x^2 + 5x – 3 &= x^2 + 4x +3\
2x^2 + 5x -3 -x^2 -4x -3 &= 0\
x^2 + x -6 &=0
end{align*}
$$

The solutions of a square equation $ax^2 + bx + c =0$ are shape

$$
x= frac {-b pm sqrt {b^2 – 4ac}}{2a}
$$

Now we have

$$
begin{align*}
x &= frac {-1 pm sqrt {1+24}}{2}\
x &= frac {-1 pm sqrt {25}}{2}\
x &= frac {-1 pm 5}{2}\
x &= – 3 qquad x=2
end{align*}
$$

If $x=-3$ we get

$$
y=0
$$

Then

$$
(x,y) = (-3,0)
$$

If $x=2$ we get

$$
y=15
$$

Then

$$
(x,y)=(2,15)
$$

$$
begin{align*}
f(x) &= 2x^2 + 5x -3- text{ blue}\
g(x) &= x^2 + 4x +3-text{ green}\
h(x) &=x^2+x-6-text{ red}
end{align*}
$$

b) Therefore

$$
begin{align*}
f(x) &> g(x)\
2x^2 + 5x – 3 &> x^2 + 4x + 3\
2x^2 + 5x – 3 &- x^2 – 4x – 3 > 0\
x^2 + x – 6 &> 0\
x2
end{align*}
$$

a) $(x,y)=(2,15)$

$(x,y) = (-3,0)$

b) $x2$

$$
7=4.2^x
$$

$ln 7=ln 4.2^x$

$$
ln 7=xln 4.2
$$

$x=dfrac{ln 7}{ln 4.2}$

$$
xapprox 1.356
$$

We divide by $ln 4.2$:

$7stackrel{?}{=}4.2^{1.356}$

$7stackrel{?}{=}7.0004513$

$$
7approx7checkmark
$$

$$
3x^5=126
$$

$x^5=dfrac{126}{3}$

$x^5=42$

$x=sqrt[5]{42}$

$$
xapprox 2.112
$$

$3(2.112)^5stackrel{?}{=}126$

$3(42.0213)stackrel{?}{=}126$

$$
126approx 126checkmark
$$

$$
14=2cdot 4^x-10
$$

$14+10=2cdot 4^x$

$24=2cdot 4^x$

$$
12=4^x
$$

$ln 12=ln 4^x$

$$
ln 12=xln 4
$$

$x=dfrac{ln 12}{ln 4}$

$$
xapprox 1.792
$$

We divide by $ln 4$:

$14stackrel{?}{=}2cdot 4^{1.792}-10$

$14stackrel{?}{=}2cdot 12-10$

$$
14=14checkmark
$$

a) $xapprox 1.356$

b) $xapprox 2.112$

c) $xapprox 1.792$

$cos AOB=dfrac{OB}{OA}=dfrac{53}{100}=0.53$

$cos^{-1} 0.53approx 58text{textdegree}$

$$
AOB=58text{textdegree}
$$

There are 4 possibilities: $A, C, E, F$. First we determine the angle $AOB$, using cosine:

$COB=180text{textdegree}-58text{textdegree}=122text{textdegree}$

$EOB=180text{textdegree}+58text{textdegree}=238text{textdegree}$

$$
FOB=360text{textdegree}-58text{textdegree}=302text{textdegree}
$$

Then we determine all the 3 angles in the other quadrants which have the same reference angle of $58text{textdegree}$:

$$
58text{textdegree}; 122text{textdegree}; 238text{textdegree}; 302text{textdegree}
$$

b) An angle with sine -1 is $270text{textdegree}$:

d) ) We sketch an angle with $cos thetaapprox -0.9,sinthetaapprox 0.4$: it lies in Quadrant II:

$BM^2+OM^2stackrel{?}{=}OB^2$

$sin^2theta+cos^2thetastackrel{?}{=}1$

$0.9^2+0.8^2stackrel{?}{=}1$

$$
1.45not=1
$$

e) We check if we can have:

$sintheta=0.9$

$costheta=0.8$.

$theta=70text{textdegree}$

$R(x,y)$

$sin theta=dfrac{y}{OR}$

$y=ORsintheta=1cdotsin 70text{textdegree}approx 0.9397$

$cos theta=dfrac{x}{OR}$

$x=ORcostheta=1cdotcos 70text{textdegree}approx 0.3420$

$$
R(0.3420,0.9397)
$$

a) We approximate the coordinates of the angle $theta$ using sine and cosine:

$$
R(cos 70text{textdegree},sin 70text{textdegree})
$$

b) The exact coordinates of the point $R$ are:

$$
sin^2theta+cos^2theta=0.9397^2+0.3420^2approx 1checkmark
$$

c) We check that the Pythagorean Identity works for the angle $theta$:

$y=sintheta$

$y=costheta$

$sin 0text{textdegree}=0$

$$
cos 0text{textdegree}=1
$$

Both functions have period $2pi$, thus we cannot use this attribute to label the functions.

We know:

Function A: $y=sin theta$

Function B: $y=cos theta$

The function which passes through $(0,0)$ is the sine function, while the function which passes through $(0,1)$ is the cosine function.

So we have:

A: $y=sin theta$

B: $y=cos theta$

$$
begin{cases}
y=cos x\
y=-1
end{cases}
$$

$cos x=-1$

$x=-3pi$

$x=-pi$

$x=pi$

$x=3pi$

$$
x=5pi
$$

$x=(2n+1)pi, n$ integer

We note on the graph that $x=-2$ is a double root and $x=2$ is a single root, thus the function is then of the form:

$$
y=a(x+2)^2(x-2)
$$

We also note that if $x=0$, then we need $y=2$:

$$
2=y=a(2)^2(-2)=-8a
$$

Divide both sides of the equation by $-8$:

$$
-dfrac{1}{4}=a
$$

Thus we then obtain the equation:

$$
y=-dfrac{1}{4}(x+2)^2(x-2)
$$

$$
y=-dfrac{1}{4}(x+2)^2(x-2)
$$

$$
f(x)=dfrac{sqrt{x+4}}{2}-1
$$

a) In order to graph $f(x)$ we start with the parent function $f_0(x)=sqrt x$. We shift $f_0$ 4 units to the left to get $f_1(x)=sqrt{x+4}$, then we vertically shrink $f_1$ by a factor of $dfrac{1}{2}$ to get $f_2(x)=dfrac{sqrt{x+4}}{2}$ and finally we shift $f_2$ 1 unit down to get $f(x)$:

We graph the inverse by reflecting $f(x)$ across the line $y=x$:

$y=dfrac{sqrt{x+4}}{2}-1$

$x=dfrac{sqrt{y+4}}{2}-1$

$x+1=dfrac{sqrt{y+4}}{2}$

$2(x+1)=sqrt{y+4}$

$[2(x+1)]^2=(sqrt{y+4})^2$

$4(x+1)^2=y+4$

$4x^2+8x+4-4=y$

$y=4x^2+8x$

$f^{-1}(x)=4x^2+8x$

$$
[-1,infty)
$$

$$
[-4,infty)
$$

The word “radian” comes from radius: it is the measure of a central angle subtending an arc equal in length to the radius.

A radian equals to $dfrac{360text{textdegree}}{2pi}approx 57.2958text{textdegree}$.

Let us wrap the radii around the circle (each $r$ represents the angle of $1$ radian):

To construct an angle that measures $1$ radian, we wrapped the radius of the circle around the circle and connected its end points with the center.

$$
r=1
$$

$$
A=pi r^2=pi(1^2)=pi
$$

$$
L=2pi r=2pi(1)=2pi
$$

$$
dfrac{L}{r}=dfrac{2pi}{1}=2piapprox 6.28
$$

$$
dfrac{L}{r}=dfrac{2pi r}{r}=2piapprox 6.28
$$

c) We determine the number of radii needed to wrap completely around the circle when the radius is $r$:

$$
2pi
$$

d) We determine the number of radians in $360text{textdegree}$: as $360text{textdegree}$ cover the entire circle, the number of radians is:

$$
dfrac{2pi}{4}=dfrac{pi}{2}
$$

We determine the number of radians in $90text{textdegree}$: as $90text{textdegree}$ cover a quarter of a circle, the number of radians is:

$$
theta=1text{textdegree}
$$

$theta=1$ radian

$theta=pi$ radians

$theta=dfrac{pi}{2}$ radians

$theta=dfrac{pi}{4}$ radians

$theta=dfrac{pi}{3}$ radians

$theta=dfrac{pi}{6}$ radians

According to the calculator, the following values of radians and degrees are equal:
$$begin{align*}
30^{circ}&=0.523text{ rad}\
60^{circ}&=1.047text{ rad}\
90^{circ}&=1.57text{ rad}\
end{align*}$$

$$begin{align*}
120^{circ}&=2.09text{ rad}\
180^{circ}&=pitext{ rad}\
270^{circ}&=4.712text{ rad}\
360^{circ}&=2pitext{ rad}
end{align*}$$

$$
sin 60text{textdegree}=0.8660254
$$

a) We determine $sin 60text{textdegree}$ with the calculator set in degree mode:

$$
sin dfrac{pi}{3}=0.8660254
$$

We determine $sin dfrac{pi}{3}$ with the calculator set in radian mode:

$$
60text{textdegree}=dfrac{pi}{3}
$$

$$
sin dfrac{pi}{4}=0.70710678
$$

b) We compute $sin dfrac{pi}{4}$:

$$
sin 45text{textdegree}=0.70710678
$$

As $dfrac{pi}{4}=45text{textdegree}$, we have:

$$
sin 135text{textdegree}=0.70710678
$$

$pi$ radians $=180$$text{textdegree}$

a-

$sin (dfrac {pi}{4})=sin 45$

$sin (dfrac {pi}{4})=dfrac {1}{sqrt {2}}=0.7071$

b-

$sin(dfrac {2pi}{3})=sin (120)$

$sin(dfrac {2pi}{3})=sin 60$          (sin (180-60)=sin 60)

$sin(dfrac {2pi}{3})=dfrac {sqrt 3}{2}=0.8660$

a-          $sin (dfrac {pi}{4})=dfrac {1}{sqrt {2}}=0.7071$

b-          $sin(dfrac {2pi}{3})=dfrac {sqrt 3}{2}=0.8660$

Colleen is calculate

$$
sin ( 30 text{radians}) = – 0.9880316241
$$

Jolleen is calculate

$$
sin ( 30^{text{textdegree}}) = 0.5
$$

Both calculators are correct.

$$
{color{#4257b2}text{a)}}
$$

Solution to this example is given below

$$
begin{align*}
log _2left(xleft(x-2right)right)&=3&&boxed{text{Apply log rule}}\
log _2left(xleft(x-2right)right)&=log _2left(8right)&&boxed{text{Apply log rule}}\
xleft(x-2right)&=8&&boxed{text{Simplify}}\
x^2-2x&=8&&boxed{text{Distributive property}}\
x^2-2x-8&=8-8&&boxed{text{Subtract 8 from both sides}}\
x^2-2x-8&=0&&boxed{text{Simplify}}\
end{align*}
$$

Solve with the quadratic formula

$$
begin{align*}
x_{1,:2}&=frac{-bpm sqrt{b^2-4ac}}{2a}&&boxed{text{ Use quadratic formula}}\
x_{1,:2}&=frac{-left(-2right)pm sqrt{left(-2right)^2-4cdot :1left(-8right)}}{2cdot :1}&&boxed{text{Substitute }1 text{ for }a, -2 text{ for } b, text{ and } -8 text{ for } c.} \
end{align*}
$$

First we solve $x_1$

$$
begin{align*}
x_1&=frac{-left(-2right)+sqrt{left(-2right)^2-4cdot :1left(-8right)}}{2cdot :1}&&boxed{text{Simplify}}\
x_1&=frac{2+sqrt{left(-2right)^2+4cdot :1cdot :8}}{2cdot :1}&&boxed{text{Remove parentheses: }-(-a)=a}\
x_1&=frac{2+sqrt{4+4cdot :1cdot :8}}{2cdot :1}&&boxed{text{Evaluate: }(-2)^2=4}\
x_1&=frac{2+sqrt{4+32}}{2cdot :1}&&boxed{text{Multiply the numbers: } 4cdot1cdot8=32}\
x_1&=frac{2+sqrt{4+32}}{2}&&boxed{text{Multiply the numbers: } 2cdot1=2}\
x_1&=frac{2+sqrt{36}}{2}&&boxed{text{Add the numbers: }4+32=36 }\
x_1&=frac{2+6}{2}&&boxed{text{Simplify}}\
x_1&=frac{8}{2}&&boxed{text{Add the numbers: } 2+6=8}\
x_1&=color{#c34632}{4} text{ True }&&boxed{text{Divide the numbers: } frac{8}{2}=4}\
end{align*}
$$

Second we solve $x_2$

$$
begin{align*}
x_2&=frac{-left(-2right)-sqrt{left(-2right)^2-4cdot :1left(-8right)}}{2cdot :1}&&boxed{text{Simplify}}\
x_2&=frac{2-sqrt{left(-2right)^2+4cdot :1cdot :8}}{2cdot :1}&&boxed{text{Remove parentheses: }-(-a)=a}\
x_2&=frac{2-sqrt{4+4cdot :1cdot :8}}{2cdot :1}&&boxed{text{Evaluate: }(-2)^2=4}\
x_2&=frac{2-sqrt{4+32}}{2cdot :1}&&boxed{text{Multiply the numbers: } 4cdot1cdot8=32}\
x_2&=frac{2-sqrt{4+32}}{2}&&boxed{text{Multiply the numbers: } 2cdot1=2}\
x_2&=frac{2-sqrt{36}}{2}&&boxed{text{Add the numbers: }4+32=36 }\
x_2&=frac{2-6}{2}&&boxed{text{Simplify}}\
x_2&=frac{-4}{2}&&boxed{text{Subtract the numbers: } 2-6=-4}\
x_2&=color{#c34632}{-2} text{ False }&&boxed{text{Divide the numbers: } frac{-4}{2}=-2}\
&boxed{{color{#c34632}x=4} }&&boxed{text{Final solution}}\
end{align*}
$$

$$
boxed{ color{#c34632} text{ }mathrm{Apply:log:rule}:quad log _cleft(aright)+log _cleft(bright)=log _cleft(abright) }
$$

$$
boxed{ color{#c34632} text{ }mathrm{Apply:log:rule}:quad :a=log _bleft(b^aright) }
$$

$$
boxed{ color{#c34632} text{ } mathrm{When:the:logs:have:the:same:base:::}log _bleft(fleft(xright)right)=log _bleft(gleft(xright)right)quad Rightarrow quad fleft(xright)=gleft(xright)}
$$

$$
{color{#4257b2}text{b)}}
$$

Solution to this example is given below

$$
begin{align*}
log _{10}left(2right)+log _{10}left(xright)-2log _{10}left(xright)&=-2&&boxed{text{Apply log rule}}\
log _{10}left(2right)-log _{10}left(xright)&=-2&&boxed{text{Simplify}}\
log _{10}left(2right)-log _{10}left(xright)-log _{10}left(2right)&=-2-log _{10}left(2right)&&boxed{text{Subtract }log_{10}(2) text{ from both sides}}\
-log _{10}left(xright)&=-2-log _{10}left(2right)&&boxed{text{Simplify}}\
frac{-log _{10}left(xright)}{-1}&=-frac{2}{-1}-frac{log _{10}left(2right)}{-1}&&boxed{text{Divide both sides by }-1}\
log _{10}left(xright)&=2+log _{10}left(2right)&&boxed{text{Simplify}}\
log _{10}left(xright)&=log _{10}left(200right)&&boxed{text{Apply log rule}}\
x&=200text{ True}&&boxed{text{Simplify}}\\
&boxed{{color{#c34632}x=200} }&&boxed{text{Final solution}}\
end{align*}
$$

$$
boxed{ color{#c34632} text{ }mathrm{Apply:log:rule}:quad log _aleft(cx^bright)=log _aleft(cright)+bcdot log _aleft(xright) }
$$

$$
boxed{ color{#c34632} text{ }mathrm{Apply:log:rule}:quad :a=log _bleft(b^aright) }
$$

$$
boxed{ color{#c34632} text{ } mathrm{When:the:logs:have:the:same:base:::}log _bleft(fleft(xright)right)=log _bleft(gleft(xright)right)quad Rightarrow quad fleft(xright)=gleft(xright)}
$$

$$
color{#4257b2} text{ a) }x=4
$$

$$
color{#4257b2} text{ b) }x=200
$$

a-

$4x^2-9y^2$          (Write the equation)

$(2x)^2-(3y)^2$

$=(2x+3y)(2x-3y)$          (Difference of two squares)

b-

$8x^3-2x^7$          (Write the equation)

$=2x^3(4-x^4)$          (Factor out $2x^3$)

$=2x^3(2+x^2)(2-x^2)$          (Difference of two squares)

c-

$x^4-81y^4$          (Write the equation)

$(x^2)^2-(9y^2)^2$

$=(x^2+9y^2)(x^2-9y^2)$          (Difference of two squares)

$=(x^2+9y^2)(x+3y)(x-3y)$          (Difference of two squares)

d-

$8x^3+2x^7$          (Write the equation)

$=2x^3(4+x^4)$          (Factor out $2x^3$)

a-          $4x^2-9y^2=(2x+3y)(2x-3y)$

b-          $8x^3-2x^7=2x^3(2+x^2)(2-x^2)$

c-          $x^4-81y^4=(x^2+9y^2)(x+3y)(x-3y)$

d-          $8x^3+2x^7=2x^3(4+x^4)$

a) The solution of equation:

$$
begin{align*}
frac 3x &= frac {5}{x-7}\
3(x-7) &= 5x\
3x – 21 &= 5x\
2x &= -21\
x&= – frac {21}{2}\
x &= 10.5
end{align*}
$$

The solution of equation is

$$
x = 10.5
$$

b) The solution of equation:

$$
begin{align*}
frac {2x+3}{4} – frac {x-7}{6} &= frac {2x-3}{12}\
frac {6(2x+3)-4(x-7)}{24} &= frac {2x-3}{12}\
frac {12x+18-4x+28}{24} &= frac {2x-3}{12}\
frac {8x+46}{24} &= frac {2x-3}{12}
end{align*}
$$

Multiply both sides by $24$

$$
begin{align*}
24 cdot frac {8x+46}{24} &= 24 cdot frac {2x-3}{12}\
8x+46 &= 2 cdot (2x-3)\
8x+46 &= 4x-6\
4x &= -52\
x &= -13
end{align*}
$$

The solution of equation is

$$
x=-13
$$

a) $x = 10.5$

b) $x=-13$

1 $cm^3=7.27$ g

$h=12$ cm

$d=3$ cm

$V<80$g

$V=pileft(dfrac{d}{2}right)^2h=pileft(dfrac{3}{2}right)^2cdot 12$

$=picdot 1.5^2cdot 12=27piapprox 84.82$ $cm^3$

$84.82cdot 7.27=616.6$ g

As $616>80$, the treetop is not solid.

$textcolor{#4257b2}{theta=0}$

$sin 0=0$

$cos 0=1$

$textcolor{#4257b2}{theta=dfrac{pi}{2}}$

$sin dfrac{pi}{2}=1$

$cos dfrac{pi}{2}=0$

$textcolor{#4257b2}{theta=pi}$

$sin pi=0$

$cos pi=-1$

$textcolor{#4257b2}{theta=dfrac{3pi}{2}}$

$sin dfrac{3pi}{2}=-1$

$cos dfrac{3pi}{2}=0$

$$
left{0, dfrac{pi}{2},pi,dfrac{3pi}{2}right}
$$

a) We draw the point on the unit circle corresponding to $dfrac{pi}{12}$:

$sindfrac{pi}{12}approx 0.25881905$

$$
cosdfrac{pi}{12}approx 0.96592583
$$

$$
P(0.96592583,0.25881905)approx (0.97,0.26)
$$

The coordinates of the point $P$ are:

b) We draw the points:

$Q$ – corresponding to $dfrac{11pi}{12}$

$R$ – corresponding to $dfrac{13pi}{12}$

$S$ – corresponding to $dfrac{7pi}{12}$

$$
Q(-0.97,0.26)
$$

i) We notice that the point $Q$ has the same $y$-coordinate as $P$ and the opposite $x$-coordinate:

$$
R(-0.97,-0.26)
$$

ii) We notice that the point $R$ has the same $x$-coordinate as $Q$ and the opposite $y$-coordinate:

$$
S(-0.26,0.97)
$$

iii) We notice that the angle formed by $OS$ with the $y$-axis is congruent with angle formed by $OQ$ with the $x$-axis, therefore the $x$-coordinate of $S$ is equal to the opposite of the $y$-coordinate of $Q$,while the $y$-coordinate of $S$ is equal to the opposite $x$-coordinate of $Q$:

a-

$(sin alpha)^2+(cos alpha)^2=1$

$(sin alpha)^2=1-left(dfrac {8}{17} right)^2$

$(sin alpha)^2=dfrac {289}{289}-dfrac {64}{289}$

$(sin alpha)^2=dfrac {225}{289}$

$$
sin alpha=dfrac {15}{17}
$$

b-

$angle (pi+alpha)$ lies in the third quadrant. So:

$sin (pi+alpha)=-sin alpha$

$sin (pi+alpha)=-dfrac {15}{17}$

c-

$angle (2 pi -alpha)$ lies in the fourth quadrant. So:

$cos (2 pi -alpha)=cos alpha$

$cos (2 pi -alpha)=dfrac {8}{17}$

a-          $sin alpha=dfrac {15}{17}$

b-          $sin (pi+alpha)=-dfrac {15}{17}$

c-          $cos (2 pi -alpha)=dfrac {8}{17}$

$$
theta=330text{textdegree}
$$

$$
360text{textdegree}-theta=360text{textdegree}-330text{textdegree}=30text{textdegree}
$$

$$
theta=120text{textdegree}
$$

$$
180text{textdegree}-theta=text{textdegree}-120text{textdegree}=60text{textdegree}
$$

$$
theta=113text{textdegree}
$$

$$
180text{textdegree}-theta=180text{textdegree}-113text{textdegree}=67text{textdegree}
$$

$$
theta=203text{textdegree}
$$

$$
theta-180text{textdegree}=203text{textdegree}-180text{textdegree}=23text{textdegree}
$$

a) $30text{textdegree}$

b) $60text{textdegree}$

c) $67text{textdegree}$

d) $23text{textdegree}$

a-

$sin (4)=sin left(4 times dfrac {180}{pi} right)$

$sin (4)=sin (229.18)$

$$
sin (4) approx -0.7568
$$

b-

$sin (dfrac {4 pi}{3})=sin (dfrac {4 times 180}{3})$

$sin (dfrac {4 pi}{3})=sin (240)$

$angle 240=angle (180+60)$          ($angle 240$$text{textdegree}$ $textrm { ~lies in the third quadrant}$)

$sin (dfrac {4 pi}{3}) =-sin (60)$

$$
sin (dfrac {4 pi}{3})=-dfrac {sqrt {3}}{2}
$$

$$
sin (dfrac {4 pi}{3}) approx -0.866
$$

a-          $sin (4) approx -0.7568$

b-          $sin (dfrac {4 pi}{3})=-dfrac {sqrt {3}}{2} approx -0.866$

$$
sin theta=0.5
$$

$theta_1=dfrac{pi}{6}$

$theta_2=dfrac{5pi}{6}$

The solutions in $[0,2pi)$ are:

$theta_1=dfrac{pi}{6}+2kpi$

$theta_2=dfrac{5pi}{6}+2kpi, k$ integer

$theta_1=dfrac{pi}{6}+2kpi$

$theta_2=dfrac{5pi}{6}+2kpi, k$ integer

We complete the table:

|Degrees |Radians |
|–|–|
|0$degree$ |0 |
|30$degree$ | $dfrac{pi}{6}$|
|45$degree$ |$dfrac{pi}{4}$ |
| 60$degree$|$dfrac{pi}{3}$ |
|90$degree$ |$dfrac{pi}{2}$ |
|120$degree$ |$dfrac{2pi}{3}$ |
|135$degree$ | $dfrac{3pi}{4}$|
|150$degree$ | $dfrac{5pi}{6}$|
| 180$degree$|$pi$ |
| 210$degree$|$dfrac{7pi}{6}$ |
|225$degree$ | $dfrac{5pi}{4}$|
|240 $degree$|$dfrac{4pi}{3}$ |
|270 $degree$| $dfrac{3pi}{2}$|
|300 $degree$| $dfrac{5pi}{3}$|
|315$degree$ | $dfrac{7pi}{4}$|
|330 $degree$| $dfrac{11pi}{12}$|

$$
6cdotdfrac{180text{textdegree}}{pi}approx 344text{textdegree}
$$

$$
sin 6approx -0.27
$$

b) We approximate the sine of the angle of 6 radians estimating the $y$-coordinate of the point from the diagram:

$f(0)=80,000$

$$
f(2)=2,400,000
$$

We are given (we consider year 1866 as $t=0$):

$$
f(t)=ab^t+k
$$

b) The equation showing the number of rabbits $t$ years after 1866 is:

$y=79979(5.47792)^{t}+21$

$f(5)=ab^5+k=79979(5.47792)^5+21$

$$
approx 394,000,000
$$

c) We have to determine $f(1871-1866)=f(5)$

$79,979(5.47792)^t+21=2$

$(5.47792)^t=dfrac{2}{79,979}$

$ln (5.47792)^t=ln dfrac{2}{79,979}$

$tln 5.47792=(5.47792)^t=ln dfrac{2}{79,979}$

$t=dfrac{ln dfrac{2}{79,979}}{ln 5.47792}$

$$
tapprox -6
$$

d) We determine the value of $t$ for which $f(t)=2$:

$t=-6$ corresponds to the year $1866-6=1860$.

e) As 24 rabbits were actually introduced in 1859, the model is pretty close.

The model shows a very fast grow, which is not real for a great value of $t$.

We can simplify this expression

$$
begin{align*}
frac {2x^5-6x^4-2x^2+7x-4}{x-3} &= frac {2x^4(x-3)-2x^2+6x+x-4}{x-3}\
&= frac {2x^4(x-3)-2x(x-3)+x-3-1}{x-3}\
&= frac {2x^4(x-3)-2x(x-3) + (x-3) -1}{x-3}\
&= frac {2x^4(x-3)}{x-3} – frac {2x(x-3)}{x-3} + frac {x-3}{x-3} – frac {1}{x-3}\
&= 2x^4 – 2x + 1 – frac {1}{x-3}
end{align*}
$$

The expression is

$$
frac {2x^5-6x^4-2x^2+7x-4}{x-3} = 2x^4 – 2x + 1 – frac {1}{x-3}
$$

$$
frac {2x^5-6x^4-2x^2+7x-4}{x-3} = 2x^4 – 2x + 1 – frac {1}{x-3}
$$

$m_{OB}=dfrac{y_B-y_O}{x_B-x_O}$

$$
=dfrac{sin AOB}{cos AOB}=dfrac{sin dfrac{pi}{10}}{cos dfrac{pi}{10}}
$$

We determine the slope of the line $OB$ in terms of $sine$ and $cosine$:

$m_{OB}=dfrac{sin dfrac{pi}{10}}{cos dfrac{pi}{10}}$

$=tan dfrac{pi}{10}$

$$
=tandfrac{pi}{10}
$$

b) We express the slope in terms of $tagent$:

We graph the function $y=tan x=dfrac{sin x}{cos x}$:

$x=dfrac{(2k+1)pi}{2}, k$ integer

The domain: $mathbb{R}-left{dfrac{(2k+1)pi}{2}, k text{ integer}right}$

The tangent function is undefined for the values of $x$ for which $cos x=0$, which must be excluded from the function’s domain:.

$$
T=pi
$$

The tangent is similar to the $sine$ and $cosine$ functions as it has a period $T$, but while the period of the functions $sine$ and $cosine$ is $2pi$, the period to the $tangent$ function is:

$x=dfrac{(2k+1)pi}{2}, k$ integer

The $sine$ and $cosine$ functions have the range $[-1,1]$, while the range of the $tangent$ function is $mathbb{R}$.

The $sine$ and $cosine$ functions increase and decrease, while the $tangent$ is only increasing.

The $sine$ and $cosine$ functions do not have asymptotes, while the $tangent$ function has vertical asymptotes:

begin{center}
begin{tabular}{|| c|c|c|c| c||}
hline
$theta$ & $cos (theta)$ & $sin(theta)$ & $tan(theta)$ & $tan(theta)$ \
& & & & (approximate to nearest 0.01)\[0.5ex]
hline
$0textdegree$ & 0& 1& 0& 0 \
hline
$30textdegree$ &$dfrac{sqrt 3}{2}$ & 0.5 & $dfrac{sqrt 3}{3}$ & 0.58 \
hline
$45textdegree$ & $dfrac{sqrt 2}{2}$ &$dfrac{sqrt 2}{2}$ &1 &1 \
hline
$60textdegree$ & 0.5 &$dfrac{sqrt 3}{2}$ & $sqrt 3$ & 1.73 \
hline
$90textdegree$ & 0& 1& undefined& undefined\
hline
$120textdegree$ &-0.5 &$dfrac{sqrt 3}{2}$ &$-sqrt 3$ &-1.73 \
hline
$135textdegree$ & $-dfrac{sqrt 2}{2}$ & $dfrac{sqrt 2}{2}$&-1 &-1 \
hline
$150textdegree$ &$-dfrac{sqrt 3}{2}$ &0.5 & $-dfrac{sqrt 3}{3}$ & -0.58 \
hline
$180textdegree$ &-1 &0 &0 &0 \
hline
$210textdegree$ &$-dfrac{sqrt 3}{2}$ &-0.5 &$dfrac{sqrt 3}{3}$ & 0.58 \
hline
$225textdegree$ & $-dfrac{sqrt 2}{2}$ &$-dfrac{sqrt 3}{2}$ &1 &1 \
hline
$240textdegree$ & -0.5& $-dfrac{sqrt 3}{2}$& $sqrt 3$& 1.73 \
hline
$270textdegree$ & 0&-1 &undefined &undefined \
hline
$300textdegree$ &0.5 &$-dfrac{sqrt 3}{2}$ & $-sqrt 3$& -1.73 \
hline
$330textdegree$ &$dfrac{sqrt 3}{2}$ &-0.5 & $-dfrac{sqrt 3}{3}$ & -0.58 \
hline
$315textdegree$ & $dfrac{sqrt 2}{2}$ &$-dfrac{sqrt 2}{2}$ &-1 &-1 \
hline
$360textdegree$ & 1& 0&0 & 0\[1ex]
hline
end{tabular}
end{center}

$$
y=tan(theta)
$$

$$
mathbb{R}-left{dfrac{(2k+1)pi}{2}| k text { integer}right}
$$

$$
mathbb{R}
$$

$$
x=dfrac{(2k+1)pi}{2}, k text { integer}
$$

b) The tangent function has the vertical asymptotes (the points in which $cos x=0$):

$$
tan(-x)=dfrac{sin(-x)}{cos(-x)}=dfrac{-sin x}{cos x}=-tan x
$$

d) The graph of $tangent$ has a different period $(pi$) than the period of $sine$ and $cosine$ ($2pi$).

The graph of $tangent$ has vertical asymptotes, while the graphs of $sine$ and $cosine$ do not.

The graph of $tangent$ increases, while the $sine$ and $cosine$ increase and decrease.

a. The domain of the function is:

$$
mathbb{R}-left{dfrac{(2k+1)pi}{2}| k text { integer}right}
$$

Please see solution for parts b-d.

a) We draw the point corresponding to $theta=dfrac{pi}{6}$ on the unit circle:

$cos BOC=dfrac{OC}{OB}$

$x_B=OC=OBcos dfrac{pi}{6}=1cdot dfrac{sqrt 3}{2}=dfrac{sqrt 3}{2}$

$sin BOC=dfrac{BC}{OB}$

$y_B=BC=OBsin dfrac{pi}{6}=1cdot dfrac{1}{2}=dfrac{1}{2}$

$$
Rightarrow Bleft(dfrac{sqrt 3}{2},dfrac{1}{2}right)
$$

We determine the coordinates $(x_B,y_B)$ of point $B$:

b) i) We draw the point corresponding to $theta=dfrac{7pi}{6}$ on the unit circle:

$cos DOE=dfrac{OE}{OD}$

$OE=ODcos dfrac{pi}{6}=1cdot left(dfrac{sqrt 3}{2}right)=dfrac{sqrt 3}{2}$

$x_D=cos AOD=-dfrac{sqrt 3}{2}$

$sin DOE=dfrac{DE}{OD}$

$DE=ODsin dfrac{pi}{6}=1cdot dfrac{1}{2}=dfrac{1}{2}$

$y_D=sin AOD=-dfrac{1}{2}$

$$
tandfrac{7pi}{6}=dfrac{sin AOD}{cos AOD}=dfrac{y_D}{x_D}=dfrac{-dfrac{1}{2}}{-dfrac{sqrt 3}{2}}=dfrac{1}{sqrt 3}=dfrac{sqrt 3}{3}
$$

We determine the coordinates $(x_D,y_D)$ of point $D$ and compute $tandfrac{7pi}{6}$:

ii) We draw the point corresponding to $theta=dfrac{13pi}{6}$ on the unit circle:

$$
cos dfrac{13pi}{6}=cosdfrac{pi}{6}=dfrac{sqrt 3}{2}
$$

As the cosine function has the period $2pi$,computing $cos dfrac{13pi}{6}$ is the same thing as computing $cos left(2pi+dfrac{pi}{6}right)=cos dfrac{pi}{6}$. We did this in part $b(i)$:

iii) We draw the point corresponding to $theta=dfrac{2pi}{3}$ on the unit circle:

$cos HOK=dfrac{OK}{OH}$

$OK=OHcos dfrac{pi}{3}=1cdot left(dfrac{1}{2}right)=dfrac{1}{2}$

$x_H=cos AOH=-dfrac{1}{2}$

$sin HOK=dfrac{HK}{OH}$

$HK=OHsin dfrac{pi}{3}=1cdot dfrac{sqrt 3}{2}=dfrac{sqrt 3}{2}$

$y_H=sin AOH=dfrac{sqrt 3}{2}$

$$
tandfrac{2pi}{3}=dfrac{sin AOH}{cos AOH}=dfrac{y_H}{x_H}=dfrac{dfrac{sqrt 3}{2}}{-dfrac{1}{2}}=-sqrt 3
$$

We determine the coordinates $(x_H,y_H)$ of point $H$ and compute $tandfrac{2pi}{3}$:

$$
theta=dfrac{7pi}{3}
$$

$$
theta=dfrac{7pi}{3}cdot dfrac{180text{textdegree}}{pi}=420text{textdegree}
$$

a) Any angle in the shape:

$alpha=theta+kcdot 360text{textdegree}=420text{textdegree}+kcdot 360text{textdegree}$ integer

corresponds to the same point on the circle.

$$
k=-1Rightarrowalpha=420text{textdegree}+(-1)360text{textdegree}=60text{textdegree}
$$

$sin AOB=dfrac{BC}{OB}$

$BC=OBsin 60text{textdegree}=1cdot dfrac{sqrt 3}{2}=dfrac{sqrt 3}{2}$

$cos AOB=dfrac{OC}{OB}$

$OC=OBcos 60text{textdegree}=1cdot dfrac{1}{2}=dfrac{1}{2}$

c) We determine the coordinates of the point $B$:

$sin dfrac{7pi}{3}=sin dfrac{pi}{3}=dfrac{sqrt 3}{2}$

$cos dfrac{7pi}{3}=cos dfrac{pi}{3}=dfrac{1}{2}$

$tan dfrac{7pi}{3}=tan dfrac{pi}{3}=dfrac{sin dfrac{pi}{3}}{cos dfrac{pi}{3}}$

$$
=dfrac{dfrac{sqrt 3}{2}}{dfrac{1}{2}}=sqrt 3
$$

We determine $sin dfrac{7pi}{3}, cos dfrac{7pi}{3}, tan dfrac{7pi}{3}$:

a) $sin180text{textdegree}=0$

b) $sin360text{textdegree}=sin (360text{textdegree}-360text{textdegree})=sin 0text{textdegree}=0$

c) $sin(-90text{textdegree})=sin (-90text{textdegree}+360text{textdegree})=sin 270text{textdegree}=-1$

d) $sin510text{textdegree}=sin (510text{textdegree}-360text{textdegree})$

$=sin 150text{textdegree}=sin (180text{textdegree}-150text{textdegree})$

$=sin 30text{textdegree}=0.5$

e) $cos 90text{textdegree}=0$

f) $tan (-90text{textdegree})$ (undefined!)

$2pi$ radians……………….$360text{textdegree}$

$x$ radians………. $ytext{textdegree}$

$$
y=dfrac{360text{textdegree}}{2pi}cdot x=dfrac{180text{textdegree}}{pi}cdot x
$$

$2pi$ radians correspond to $360text{textdegree}$. Then $x$ radians correspond to:

$360text{textdegree}$……………$2pi$ radians

$xtext{textdegree}$………. $y$ radians

$$
y=dfrac{2pi}{360text{textdegree}}cdot x=dfrac{pi}{180text{textdegree}}cdot x
$$

$360text{textdegree}$ correspond to $2pi$ radians. Then $xtext{textdegree}$ correspond to:

$$
dfrac{7pi}{6}=dfrac{7pi}{6}cdot dfrac{180text{textdegree}}{pi}=210text{textdegree}
$$

$$
dfrac{5pi}{3}=dfrac{5pi}{3}cdot dfrac{180text{textdegree}}{pi}=300text{textdegree}
$$

$$
45text{textdegree}=45text{textdegree}cdotdfrac{pi}{180text{textdegree}}=dfrac{pi}{4}
$$

$$
100text{textdegree}=100text{textdegree}cdotdfrac{pi}{180text{textdegree}}=dfrac{5pi}{9}
$$

$$
810text{textdegree}=810text{textdegree}cdotdfrac{pi}{180text{textdegree}}=dfrac{9pi}{2}
$$

$$
dfrac{7pi}{2}=dfrac{7pi}{2}cdot dfrac{180text{textdegree}}{pi}=630text{textdegree}
$$

a) $210text{textdegree}$; b) $300text{textdegree}$; c) $dfrac{pi}{4}$; d) $dfrac{5pi}{9}$; e) $dfrac{9pi}{2}$; f) $630text{textdegree}$

$costheta=-dfrac{12}{13}$ (Quadrant III)

$sin^2theta+left(-dfrac{12}{13}right)^2=1$

$sin^2theta+dfrac{144}{169}=1$

$sin^2theta=1-dfrac{144}{169}$

$sin^2theta=dfrac{25}{169}$

$sintheta=pmdfrac{5}{13}$

a)We use the Pythagorean identity:

$sin^2theta+cos^2theta=1$.

$sintheta=-dfrac{5}{13}$

Because $theta$ is in Quadrant III, the sine is negative:

$$
tantheta=dfrac{sintheta}{costheta}=dfrac{-dfrac{5}{13}}{-dfrac{12}{13}}=dfrac{5}{12}
$$

b) We determine $tantheta$:

a) $sintheta=-dfrac{5}{13}$

b) $tantheta=dfrac{5}{12}$

The solution of equation:

$$
log_2(x) = 2^x
$$

Let

$$
begin{align*}
y &= log_2(x)-text{blue}\
y &= 2^x-text{red}
end{align*}
$$

If $a,b$ and $c$ are solutions of the equation, then the equation is of the form

$$
color{#c34632}{ (x-a)(x-b)(x-c) =0}
$$

Since

$$
begin{align*}
a&=3\
b&=2\
c&=-1
end{align*}
$$

We obtain

$$
begin{align*}
(x-3)(x-2)(x+1) &= 0\
(x^2 -2x-3x+6)(x+1) &= 0\
(x^2 – 5x + 6)(x+1) &= 0\
x^3 – 5x^2 + 6x + x^2 -5x +6 &= 0\
x^3 – 4x^2 + x +6 &= 0
end{align*}
$$

In order for the function to pass through point $(1,1)$, it is necessery to calculation $k$ such that

$$
k (x^3 -4x^2 + x +6) = 0
$$

Since $(1,1)$ we obitan

$$
y(1) =1
$$

Then

$$
begin{align*}
k ( 1^3 – 4 cdot 1^2 + 1 +6) &= 1\
k (1-4+1+6) &= 1\
4k &= 1\
k &= frac 14
end{align*}
$$

The function passing through point $(1,1)$ is

$$
y = frac 14 ( x^3 – 4x^2 + x +6)
$$

$$
y = frac 14 ( x^3 – 4x^2 + x +6)
$$

a) In the unit circle, $x=costheta$, while $y=sintheta$.

b) In graph $b$, $x$ represents the angle, while $y=sin x$.

c) In graph $c$, $x$ represents the angle, while $y=cos x$.

d) In graph $d$, $x$ represents the angle, while $y=tan x$.

Let us graph $y=sin x$ and $y=cos x$.

The domain of the functions $sin x$ and $cos x$ is $(-infty,+infty)$. The range of the functions $sin x$ and $cos x$ is $(-1,1)$. They do not have any asymptotes.

Let us graph $sin x$, $sin 2x$, $2sin x$, $sin(x-1)$ and $sin x+1$:

The general form of the equation of the sine function is:
$$y=Asin(B(x-k))+c,$$
where $k$ represents horizontal shift, $c$ represents the vertical shift, $B$ represents the horizontal stretching and $|A|$ is an amplitude.

The same goes for the cosine function:
$$y=Acos(B(x-k))+c.$$

Usually $A$ represents the vertical stretching or compression (it does here too), but it also tells us more important information about the curves. It tells us the amplitude of $sin$ and $cos$.

$$
y=sin (x)
$$

$y=sin (x)+3$

i) We shift $y=sin (x)$ 3 units up:

$y=-sin (x)$

ii) We reflect $y=sin (x)$ across the $x$-axis:

$y=sin (x-2)$

iii) We shift $y=sin (x)$ 2 units to the right:

$y=2sin x$

iv) We vertically stretch $y=sin (x)$ by a factor of 2:

$$
y=asin (bx+c)+d
$$

$$
y=sin (x)
$$

$$
y=sin (x)+1
$$

b) In order to get the function $y=sin (x)$ shifted up 1 unit we have to add 1 to the function:

$…, -dfrac{pi}{2}, dfrac{3pi}{2}, dfrac{7pi}{2},….$

c) The $x$-intercepts of the function $y=sin (x)+1$ are:

$$
y=1
$$

The $y$-intercept is:

$x=dfrac{(4k+3)pi}{2}, k$ integer

d) All the $x$-intercepts are:

a) The above function was obtained from the parent function $y=sin (x)$ by shifting it by $pi$ to the left.

$$
y=sin (x-pi)
$$

$$
y = acos(x-h) + k
$$

Horizontal shift: $h=0$

Vertical shift: $k=-50$

Amplitude: $a=100$

$$
y=100cos x-50
$$

$$
y=100cos x-50
$$

$y=sin (x)$

$$
y=cos (x)
$$

$y=sin (x)=cosleft(x-dfrac{pi}{2}right)$

$y=cos (x)=sinleft(x-dfrac{pi}{2}right)$

$$
y=12x^3+55x^2-27x-10
$$

$$
polyhornerscheme[x=-5]{12x^3+55x^2-27x-10}
$$

Since $(-5,0)$is a $x$-intercept, the polynomial has the factor $x+5$. We divide the polynomial by $x+5$ using synthetic division:

$$
y=(x+5)(12x^2-5x-2)
$$

$y=(x+5)(12x^2-8x+3x-2)$

$=(x+5)[4x(3x-2)+(3x-2)]$

$$
(x+5)(3x-2)(4x+1)
$$

$3x-2=0Rightarrow 3x=2Rightarrow x=dfrac{2}{3}$

$4x+1=0Rightarrow 4x=-1Rightarrow x=-dfrac{1}{4}$

We determine the other 2 $x$-intercepts:

$$
left{-dfrac{1}{4},dfrac{2}{3}right}
$$

The angle $alpha$ can be calculate besad on the formula

$$
cos alpha = frac {frac a2}{b}
$$

Then

$$
begin{align*}
cos alpha &= frac {2.5}{10}\
cos alpha &= 0.25
end{align*}
$$

We can use the inverse function

$$
begin{align*}
alpha &= arccos 0.25\
alpha &= 75.5^{text{textdegree}}
end{align*}
$$

The sum of the interior angles of a triangle is $180^{text{textdegree}}$

$$
begin{align*}
alpha + alpha + beta &= 180^{text{textdegree}}\
75.5^{text{textdegree}} + 75.5^{text{textdegree}} + beta &= 180^{text{textdegree}}\
151^{text{textdegree}} + beta &= 180^{text{textdegree}}\
beta &= 29^{text{textdegree}}
end{align*}
$$

The angles in the triangle are

$$
75.5^{text{textdegree}} , qquad 75.5^{text{textdegree}} qquad text{and} qquad 29^{text{textdegree}}
$$

$$
75.5^{text{textdegree}} , qquad 75.5^{text{textdegree}} qquad text{and} qquad 29^{text{textdegree}}
$$

$$
y=asin(x-h)+k
$$

$$
y=asin (bx-h)+k
$$

The general equation does not include all the possible transformations of the graph of $y=sin (x)$.

For example:

The following table represents the distance Briana’s radar.travels through time.\
begin {center}
renewcommand{arraystretch}{2.5}
begin{tabular}{ | m{2cm} | m{2cm} || m{2cm} | m{2cm} | }
hline
t (seconds) & d(t) & t (seconds) & d(t) \
hline
1 & $dfrac { pi}{12}$ & 13 & $dfrac {13 pi}{12}$ \
hline
2 & $dfrac { pi}{6}$ & 14 & $dfrac {7 pi}{6}$ \
hline
3 & $dfrac { pi}{4}$ & 15 & $dfrac {5 pi}{4}$ \
hline
4 & $dfrac { pi}{3}$ & 16 & $dfrac {4 pi}{3}$ \
hline
5 & $dfrac {5 pi}{12}$ & 17 & $dfrac {17 pi}{12}$ \
hline
6 & $dfrac { pi}{2}$ & 18 & $dfrac {3 pi}{2}$ \
hline
7 & $dfrac {7 pi}{12}$ & 19 & $dfrac {19 pi}{12}$ \
hline
8 & $dfrac {2 pi}{3}$ & 20 & $dfrac {5 pi}{3}$ \
hline
9 & $dfrac {3 pi}{4}$ & 21 & $dfrac {7 pi}{4}$ \
hline
10 & $dfrac {5 pi}{6}$ & 22 & $dfrac {11 pi}{6}$ \
hline
11 & $dfrac {11 pi}{12}$ & 23 & $dfrac {23 pi}{12}$ \
hline
12 & $pi$ & 24 & $2 pi$ \
hline
end {tabular}
end{center}

Since it takes 2 seconds for the radar line to travels through an angle of $dfrac {pi}{6}$

Then, we can calculate the output with the equation:

$d(t)=dfrac {t}{2}times dfrac { pi}{6}$

It takes 24 seconds to complete on full cycle on the radar screen

From the graph, the line reaches $2 pi$ on y-axis at 24 seconds on x-axes

We can calculate the output with the equation:          $d(t)=dfrac {t}{2}times dfrac { pi}{6}$

It takes 24 seconds to complete on full cycle on the radar screen

$$
y=asin(x-h)+k
$$

a) A new perimeter can be introduced as a coefficient of $x$.

b) We graph the functions:

$y=sin (2x-1)+1$

$y=sin (3x-1)+1$

$y=sin (4x-1)+1$

As the coefficient of $x$ increases, the graph shrinks more and more.

We graph the functions:

$y=sin (0.5x-1)+1$

$y=sin (0.3x-1)+1$

$y=sin (0.1x-1)+1$

As the coefficient of $x$ decreases, the graph stretches more and more.

$$
y=asin (bx-h)+k
$$

$$
y=asin(bx-h)+k
$$

$$
T_1=pi
$$

The graph of the function repeats after $pi$, therefore its period is:

$$
T_2=2pisqrt{dfrac{L}{g}}=3
$$

b) The period T_2$ of a pendulum is:

$$
T_3=2pi
$$

c) The period $T_3$ of the function $y=sin (x)$ is:

Period of function $sin x$ is $2pi$.

$$
y=sin (x-pi)
$$

$sin x=cosleft(dfrac{pi}{2}-xright)$

$y=cosleft(dfrac{pi}{2}-x+piright)$

$y=cosleft(dfrac{3pi}{2}-xright)$

$$
y=-sin (x)
$$

$$
y=-sin x
$$

Claudia did not saw any difference between the graph of $y=cos theta$ and the graph of $y=cos (theta+360text{textdegree})$ because the period of $cos$ function is $360text{textdegree}$ or $2pi$. The value of trigonometric function does not changes for multiple of $2pi$ or $360text{textdegree}$, Hence the graph of $y=cos theta$ and the graph of $y=cos (theta+360text{textdegree})$ will overlap each other.

So the value of $y=cos theta=cos (theta+360text{textdegree})=cos (theta+720text{textdegree})=cos (theta+n2pi)$ will be same.

a) According to the formula

$$
begin{align*}
color{#c34632}{tan theta = frac {sin theta}{cos theta}} tag{$*$}
end{align*}
$$

we obtain

$$
tan left( frac {2 pi}{3} right) = frac {sin left( frac {2 pi}{3} right)}{cos left( frac {2 pi}{3} right)}
$$

Since

$$
begin{align*}
sin left( frac {2 pi}{3} right) &= frac {sqrt 3}{2}\
cos left( frac {2 pi}{3} right) &= – frac 12
end{align*}
$$

we have

$$
tan left( frac {2 pi}{3} right) = frac {frac {sqrt 3}{2}}{- frac 12} = – sqrt 3
$$

b) According to the $(*)$ we obtain

$$
tan left( frac {7 pi}{6} right) = frac {sin left( frac {7 pi}{6} right)}{cos left( frac {7 pi}{6} right)}
$$

Since

$$
begin{align*}
sin left( frac {7 pi}{6} right) &= – frac 12\
cos left( frac {7 pi}{6} right) &= – frac {sqrt 3}{2}
end{align*}
$$

we have

$$
tan left( frac {7 pi}{6} right) = frac {- frac 12}{ – frac {sqrt 3}{2}} = frac {1}{sqrt 3} = frac {sqrt 3}{3}
$$

a) $tan left( frac {2 pi}{3} right) = – sqrt 3$

b) $tan left( frac {7 pi}{6} right) = frac {sqrt 3}{3}$

$a=1$

$b=0.5$

$$
c=dfrac{sqrt 3}{2}
$$

$0.5<dfrac{sqrt 3}{2}<1$

$$
b<c<a
$$

$b^2+c^2stackrel{?}{=}a^2$

$0.5^2+left(dfrac{sqrt 3}{2}right)^2stackrel{?}{=}1^2$

$0.25+dfrac{3}{4}stackrel{?}{=}1$

$0.25+0.75stackrel{?}{=}1$

$$
1=1checkmark
$$

Because $b^2+c^2=a^2$, the triangle is a right triangle.

Compound interest equation:

$$
K=k(1+r)^n
$$

Replace $K$ with 25,000, $k$ with $15,000$ and $r$ with $8%=0.08$:

$$
25,000=15,000(1.08)^n
$$

Divide both sides of the equation by 15,000:

$$
dfrac{5}{3}=(1.08)^n
$$

Take the logarithm with base 1.08 from both sides of the equation:

$$
log_{1.08}{dfrac{5}{3}}=n
$$

Determine $n$:

$$
n=log_{1.08}{dfrac{5}{3}}=dfrac{log{dfrac{5}{3}}}{log{1.08}}approx 7
$$

Thus it will take approximately 7 years.

The polynomial $f(x)$ can be write

$$
begin{align*}
f(x) &= x^3 – 5x^2 + 11x – 15\
&= x^3 – 3x^2 -2x^2 + 6x + 5x -15\
&= x^2 (x-3) – 2x (x-3) + 5 (x-3)\
&= (x-3)( x^2 – 2x + 5)
end{align*}
$$

The possible factor is

$$
d) (x-3)
$$

d) $(x-3)$

a) The factor of this polynomial is $(x-3)$

$$
begin{align*}
f(x) &= x^3 – 5x^2 + 11x -15\
&= (x-3)(x^2 – 2x +5)
end{align*}
$$

b) The roots is

$$
begin{align*}
x-3 &= 0\
x &= 0
end{align*}
$$

and

$$
x^2 -2x + 5 =0
$$

The solution of a square equation $a^2 + bx + c =0$ are shape

$$
x = frac {-b pm sqrt {b^2 – 4ac}}{2a}
$$

Now we have

$$
begin{align*}
x &= frac {2 pm sqrt {4-20}}{2}\
x &= frac {2 pm sqrt {-16}}{2}\
x &= frac {2 pm 4 i}{2}\
x &= 1+ 2 i qquad x= 1- 2 i
end{align*}
$$

The roots of the polynomial

$$
x=3 qquad x=1+2 i qquad x=1-2 i
$$

a) $f(x) = (x-3)(x^2 – 2x +5)$

b) $x=3 qquad x=1+2 i qquad x=1-2 i$

$$
y=costheta
$$

$$
T=2pi
$$

$T=365$ days

a-

The domain of one cycle of the function $y=sin(x)$ is:          $0 leq x leq 2 pi$

The range of one cycle of the function $y=sin(x)$ is:          $-1 leq y leq 1$

b-\
The table below represents the cycles number, midline, amplitude and cycle length of each graph.\
begin {center}
renewcommand{arraystretch}{2.5}
begin{tabular}{|l|c|c|c|c|c|}
hline
Function & Cycles # & Midline Equation & Amplitude & Period & Function? (Yes/No) \
hline
$y=sin(x)$ & 1 & $y=0$ & 1 & $2 pi$ & Yes \
hline
$y=sin(0.5 x)$ & $dfrac {1}{2}$ & $y=0$ & 1 & $4 pi$ & Yes \
hline
$y=sin(2 x)$ & 2 & $y=0$ & 1 & $dfrac {1}{2} pi$ & Yes \
hline
$y=sin(3 x)$ & 3 & $y=0$ & 1 & $dfrac {2}{3} pi$ & Yes \
hline
$y=sin(5 x)$ & 5 & $y=0$ & 1 & $dfrac {2}{5} pi$ & Yes \
hline
end{tabular}
end {center}
All the graph are equations because they pass the vertical line test.\

c-

For the graph of $y=sin(bx)$

i.          The cycles number $=b$

ii.          The midline equation is: $y=0$

iii.          The amplitude: $=1$

iv.          The period (Cycle length) is: $dfrac {2 pi}{b}$

v.          The equation is a function.

d-

The graph of the equation $y=sin(6x)$

i.          The cycles number $=6$

ii.          The midline equation is: $y=0$

iii.          The amplitude: $=1$

iv.          The period (Cycle length) is: $dfrac {2 pi}{6}$

v.          The equation is a function.

The graph of the equation $y=sin(7x)$

i.          The cycles number $=7$

ii.          The midline equation is: $y=0$

iii.          The amplitude: $=1$

iv.          The period (Cycle length) is: $dfrac {2 pi}{7}$

v.          The equation is a function.

The graph of the equation $y=sin(8x)$

i.          The cycles number $=8$

ii.          The midline equation is: $y=0$

iii.          The amplitude: $=1$

iv.          The period (Cycle length) is: $dfrac {2 pi}{8}$

v.          The equation is a function.

e-

For the graph of $y=sin(bx)$, the relation between the period of the sing graph and $b$ is:

The period (Cycle length) is: $dfrac {2 pi}{b}$

For the graph of $y=sin(bx)$

i.          The cycles number $=b$

ii.          The midline equation is: $y=0$

iii.          The amplitude: $=1$

iv.          The period (Cycle length) is: $dfrac {2 pi}{b}$

v.          The equation is a function.

$$
y=sin 2left(x-dfrac{pi}{6}right)
$$

Amplitude: $A=1$

Period: $T=dfrac{2pi}{2}=pi$

Horizontal shift: $-dfrac{-dfrac{pi}{6}}{1}=dfrac{pi}{6}$

Vertical shift: $0$

$$
y=3+sin left(dfrac{1}{3}xright)
$$

Amplitude: $A=1$

Period: $T=dfrac{2pi}{dfrac{1}{3}}=6pi$

Horizontal shift: $-dfrac{0}{dfrac{1}{3}}=0$

Vertical shift: $3$

$$
y=3sin (4x)
$$

Amplitude: $A=3$

Period: $T=dfrac{2pi}{4}=dfrac{pi}{2}$

Horizontal shift: $-dfrac{0}{4}=0$

Vertical shift: $0$

$$
y=sin dfrac{1}{2}left(x+1right)
$$

Amplitude: $A=1$

Period: $T=dfrac{2pi}{dfrac{1}{2}}=4pi$

Horizontal shift: $-dfrac{dfrac{1}{2}}{dfrac{1}{2}}=-1$

Vertical shift: $0$

$$
y=-sin 3left(x-dfrac{pi}{3}right)
$$

Amplitude: $A=1$

Period: $T=dfrac{2pi}{3}$

Horizontal shift: $-dfrac{-pi}{3}=dfrac{pi}{3}$

Vertical shift: $0$

$$
y=-1+sin left(2x-dfrac{pi}{2}right)
$$

Amplitude: $A=1$

Period: $T=dfrac{2pi}{2}=pi$

Horizontal shift: $-dfrac{-dfrac{pi}{2}}{2}=dfrac{pi}{4}$

Vertical shift: $-1$

Amplitude: $A=3$

Horizontal shift: $dfrac{pi}{4}$

Midline: $y=-2$

$$
y=Asin(Bx+C)+D
$$

$A=3$

$D=-2$

$$
-dfrac{C}{B}=dfrac{pi}{4}
$$

$$
C=-dfrac{pi}{4}cdot B=-dfrac{pi}{4}cdot 2=-dfrac{pi}{2}
$$

Since both used $B=2$, we determine the correct $C$:

$y=3sin left(2x-dfrac{pi}{2}right)-2$

$$
y=3sin 2left(x-dfrac{pi}{4}right)-2
$$

b) We graph the function:

$y=3sin left(2x-dfrac{pi}{4}right)-2$

in continuous line, and the correct function:

$y=3sin left(2x-dfrac{pi}{2}right)-2$

in dotted line:

The wrong function is $dfrac{pi}{8}$ units to the left of the correct function.

$y=Asin (Bx+C)+D$, where:

$A$=the amplitude

$dfrac{2pi}{B}$=the period

$-dfrac{C}{B}$=the horizontal shift

$D$=the vertical shift (midline)

If $-dfrac{C}{B}0$, the graph is pushed to the right.

$f(x)=sin (x)$

$$
g(x)=3sinleft(dfrac{1}{2} xright)
$$

a) The graph of $g(x)$ is obtained from the graph of $f(x)$ by a horizontal stretch by the factor of $dfrac{1}{2}$ and a vertical stretch by a factor of 3.

c) The graph of $g(x)$ has an amplitude 3 times greater than $f(x)$ and a period of $4pi$ instead of $2pi$. The horizontal and vertical shifts are zero in both cases.

$$
y=sin (2pi x)
$$

$$
T=dfrac{2pi}{2pi}=1
$$

$$
T=1
$$

$$
y=sin [2(x-1)]=sin (2x-2)
$$

Shifting the function $y=sin (2x)$ 1 unit to the right is correctly done if we subtract 1 from the variable $x$:

Adel got the right equation. Ceirin’s equation $y=sin (2x-1)$ is wrong because the subtraction must be made only from $x$, not from $2x$.

When having to shift a graph 1 unit to the right, we must always see the variable $x$ as being isolated from the rest of the equation and practically replace $x$ by $(x-1)$,the brackets making sure that the subtraction applies to $x$.

a) In this exercise we can use the formula

$$
sin theta = frac ax
$$

Therefore

$$
begin{align*}
sin 30^{text{textdegree}} &= frac 2x\
frac {1}{2} &= frac 2x\
x &= 2 cdot 2\
x &= 4
end{align*}
$$

b) In this exercise we can use the formula

$$
cos theta = frac ax
$$

Therefore

$$
begin{align*}
cos 45^{text{textdegree}} &= frac 4x\
frac {sqrt 2}{2} &= frac 4x\
x sqrt 2 &= 4 cdot 2\
x sqrt 2 &= 8\
x &= frac {8}{sqrt 2}\
x &= frac {8 sqrt 2}{2}\
x &= 4 sqrt 2
end{align*}
$$

a) $x = 4$

b) $x = 4 sqrt 2$

$$
29=3^x
$$

$log (29)=log(3^x)$

$log (29)=xlog(3)$

$x=dfrac{log (29)}{log (3)}$

$$
xapprox 3.0650
$$

$3^x=dfrac{1}{29}$

$$
3^x=29^{-1}
$$

$log (3^x)=log (29^{-1})$

$xlog (3)=-log (29)$

$x=dfrac{-log (29)}{log (3)}$

$$
xapprox -3.0650
$$

$$
29=x^3
$$

$x=sqrt[3] {29}$

$$
xapprox 3.0723
$$

$$
-29=x^3
$$

$x=sqrt[3] {-29}$

$$
xapprox -3.0723
$$

$A=P(1+r)^t$

$A=10(1+0.07)^t$

$$
A=10(1.07)^t
$$

$15=10(1.07)^t$

$(1.07)^t=dfrac{15}{10}$

$(1.07)^t=1.5$

$log ((1.07)^t)=log (1.5)$

$tlog (1.07)=log (1.5)$

$t=dfrac{log (1.5)}{log (1.07)}$

$$
tapprox 6
$$

We determine the time $t$ for which $A=15$:

a) $xapprox 3.0650$

b) $xapprox -3.0650$

c) $xapprox 3.0723$

d) $xapprox -3.0723$

e) 6 years

$$
begin{cases}
y=x^2-5\
y=x+1
end{cases}
$$

$x+1=x^2-5$

$x+1-x-1=x^2-5-x-1$

$x^2-x-6=0$

$x^2+2x-3x-6=0$

$x(x+2)-3(x+2)=0$

$(x+2)(x-3)=0$

$x+2=0Rightarrow x_1=-2$

$x-3=0Rightarrow x_2=3$

We substitute the expression of $y$ from the second equation into the first and determine $x$:

$x_1=-2Rightarrow y_1=-2+1=-1$

$x_2=3Rightarrow y_2=3+1=4$

We determine $y$:

$$
{(-2,-1),(3,4)}
$$

$$
{(-2,-1),(3,4)}
$$

$$
y=x^2+4x+c
$$

$Delta=4^2-4(1)(c)=16-4c<0$

$16-4c+4c<0+4c$

$16<4c$

$4<c$

The equation has complex roots if the discriminant $Delta$ is negative:

$$
5,6,7,8,9,10,11,12
$$

Therefore for $c>4$ the equation has complex roots. In the set ${1,2,3,…,11,12}$ the numbers greater than 4 are:

$$
P(c>4)=dfrac{8}{12}=dfrac{2}{3}approx 66.66%
$$

$$
dfrac{2}{3}approx 66.66%
$$

$y=Asin (Bx+C)+D$

$y=Acos (Bx+C)+D$

$cos x=sinleft(dfrac{pi}{2}-xright)$

$sin x=cosleft(dfrac{pi}{2}-xright)$

$A$=the amplitude

$B$ which gives us the period $T=dfrac{2pi}{B}$

$C$ which gives us the horizontal shift $-dfrac{C}{B}$

$D$=the vertical shift (midline)

$$
A=2
$$